Monday, March 7, 2016

Fractions for Adults

The following is from my teaching experience. My students in college and university were at least 18 years old, often older and sometimes middle aged. And some of them have trouble with fractions. Rules for fractions are not complicated. The problem is that there are many of them, some of them similar, and you should know what particular rule in your case to choose. A concrete example would be helpful to keep in mind .

I have looked different explanations. The simple ones have been using cutting pizza or cake to pieces and were taken from elementary school textbooks. It was too childish for my students. Other presentations had a lot of letter notation and were not concrete enough.

Then I recalled an advice from a experienced instructor at the dawn of my teaching career. She said “If you students cannot calculate, try to make your question about money. Everybody can count money.” I saw an instance of it right away: a student who was stupefied by a question “what is one plus half” was immediately able to answer “1.50” to rephrased form “one ruble and a half of it”.

Here it goes.

Let us consider usual fraction operations with dimes and nickels. A fraction \(\frac{m}{n}\) means that some unit was divided in \(n\) parts and here are \(m\) such parts. There are 10 dimes in one dollar, so one dime is \(\frac{1}{10}\) of a dollar. Or, if you want to consider the amount as pennies, it is \(\frac{10}{100}\). So \(\frac{1}{10}=\frac{10}{100}\). The same way we can represent a nickel as a fraction of a dollar: \(\frac{1}{20}\). Now we can express in fractions that one dime is 2 nickels: \(\frac{1}{10}=\frac{2}{20}\). These equalities show how fractions are cancelled:

The examples help to remember that the correct way to cancel a fraction is to divide its numerator and denominator by the same number: The simplest example is when we have a number over another number. When there is a sum, things get more complicated.

Assume that you have 4 dimes and 6 nickels, which you need to sum up. We forget about pennies for a while and will be considering only dimes and nickels. You can take 8 nickels instead of 4 dimes \(\left(\frac{4}{10}=\frac{8}{20}\right)\) here and compute the number of the nickels. So at first we have 4 dimes and 6 nickels, then we replace dimes with nickels and calculate the number of nickels. You can see below how the process looks in fractions: \[ \frac{4}{10}+\frac{6}{20}=\frac{8}{20}+\frac{6}{20}=\frac{8+6}{20}=\frac{14}{20}
\] The total is 14 nickels, or 7 dimes: \(\frac{14}{20}= \frac{2(7)}{2(10)}= \frac{7}{10}\). Or in the original pile of 4 dimes and 6 nickels we can replace 6 nickels with 3 dimes \(\left(\frac{6}{20}=\frac{3}{10}\right)\) and get the final result in dimes: \[ \frac{4}{10}+\frac{6}{20}= \frac{4}{10}+\frac{3}{10}=\frac{4+3}{10}=\frac{7}{10}
\] Going from nickels to dimes requires cancelling fractions. We can do it another way. Start to add coins as nickels, them in the middle of calculations turn them into dimes:

Note that all terms in the numerator should be divided by 2 to turn them into dimes. Otherwise you are trying to sum coins of different monetary values. If you do it with only one term, it will be a mistake:

Of course we can say that `Here are 10 coins’’, but it does not tell us how much money the coins represent.

One more way of fraction cancelling is taking out a common factor for a sum: \[ \frac{4}{10}+\frac{6}{20}=\frac{8}{20}+\frac{6}{20}=
\frac{8+6}{20}=\frac{2(4+3)}{2(10)}=\frac{7}{10}
\] Now you see what I meant by different, but confusingly similar rules: in this case we cancel only one number on the top. Hope now you can figure out why!

I am sorry for different fonts, could not find out how to use LaTeX packages in R markdown.